elisabettaportioli wrote:

Bunuel Bunuel wrote:

Walkabout wrote:

How many prime numbers between 1 and 100 are factors of 7,150 ?

(A) One

(B) Two

(C) Three

(D) Four

(E) Five

Make prime factorization of 7,150 --> \(7,150=2*5^2*11*13\). So, 4 prime numbers between 1 and 100 (namely 2, 5, 11, and 13) are factors of 7,150.

Answer: D.

Bunuel,

Do you have any fast tricks to understand that 7,150 is divisible by 11 and 13?

The only silly mistake I made was to miss 11 and 13 as prime factors of 7150.

Thank you!

One of the questions above is how to recognize easily if a number is a multiple of 11 or not.

Let \(N\) be an integer \(a_m a_{m-1} ... a_2 a_1 a_0\) written in 10 base notation.

Consider the difference between \(a_0 + a_2 + a_4 + ...\), the sum of digits at even numbered positions such as 0th, 2nd, 4th, ... and \(a_1 + a_3 + a5 + ...\), the sum of digits at odd numbered positions such 1st, 3rd, 5th, ... .

If \(( a_0 + a_2 + a_4 + ... ) - ( a_1 + a_3 + a5 + ... )\) is a multiple of 11, then \(N\) must be a multiple of 11.

Here, in this question, we have ( 7 + 5 ) - ( 1 + 0 ) = 12 - 1 = 11, which is a multiple of 11.

Thus 7150 must be a multiple of 11.

For 143, ( 1 + 3 ) - 4 is 0, which is also a multiple of 11.

Thus 143 is a multiple of 11.

Are you curious about why?

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